Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=t^2-1$ and $y=t^4-2t^3$. What is the particle's velocity vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0,-1) $ (Choice B) B $(2,0)$ (Choice C) C $(2,-2) $ (Choice D) D $(0,12)$
Solution: Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=t^2-1$ and $y=t^4-2t^3$, which means its position vector is $(t^2-1, t^4-2t^3)$. We are asked to find the particle's velocity vector at $t=1$. In other words, we need to find $\vec{v}(1)$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^2-1),\dfrac{d}{dt}(t^4-2t^3)\right) \\\\ &=(2t,4t^3-6t^2) \end{aligned}$ Finding $\vec{v}(1)$ $\begin{aligned} \vec{v}({1})&=(2({1}),4({1})^3-6({1})^2) \\\\ &=(2,4-6) \\\\ &=(2,-2) \end{aligned}$ In conclusion, the particle's velocity vector at $t=1$ is $(2,-2)$.